GATE EC 2014 Set 4 — Question 59
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Communications → Digital Modulation & Detection → Error Probability
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Question
Consider a communication scheme where the binary valued signal X satisfies P{X = +1} = 0.75 and P{X = -1} = 0.25. The received signal Y = X + Z, where Z is a Gaussian random variable with zero mean and variance . The received signal Y is fed to the threshold detector. The output of the threshold detector is:.To achieve a minimum probability of error P{}, the threshold should be
Correct answer
(C) strictly negative
Solution
To minimize the probability of error, we use the Maximum A Posteriori (MAP) decision rule. We decide in favor of the hypothesis with the higher posterior probability.Decide X = +1 if .
Using Bayes' rule, this is equivalent to:
The received signal is . Given , is a Gaussian random variable.
If , is Gaussian with mean and variance . Its PDF is .
If , is Gaussian with mean and variance . Its PDF is .The prior probabilities are given as and .The decision boundary (threshold ) is found where the posterior probabilities are equal:
Taking the natural logarithm of both sides:
The threshold is the value of that satisfies this equation:
Since (variance is positive) and , the product is positive.
Therefore, the threshold is a negative value.Thus, the threshold should be strictly negative.
Using Bayes' rule, this is equivalent to:
The received signal is . Given , is a Gaussian random variable.
If , is Gaussian with mean and variance . Its PDF is .
If , is Gaussian with mean and variance . Its PDF is .The prior probabilities are given as and .The decision boundary (threshold ) is found where the posterior probabilities are equal:
Taking the natural logarithm of both sides:
The threshold is the value of that satisfies this equation:
Since (variance is positive) and , the product is positive.
Therefore, the threshold is a negative value.Thus, the threshold should be strictly negative.
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